Topic: Digital Logic
In order to get low output from the following logic circuit, the possible inputs are
[Logic circuit showing A and B as inputs to a NAND gate. The output of the NAND gate and C are inputs to an OR gate. The final output is Y]
- A. A-1, B-1, C-0
- B. A-1, B-1, C-1
- C. A-0, B-0, C-1
- D. A-0, B-0, C-0
Answer: A. A-1, B-1, C-0
Explanation: From the schematic, the boolean expression for the circuit is $Y = \overline{(A \cdot B)} + C$.
We want a low output, so $Y = 0$.
For an OR gate to output 0, both of its inputs must be 0. Thus, $\overline{(A \cdot B)} = 0$ AND $C = 0$.
$\overline{(A \cdot B)} = 0 \implies A \cdot B = 1 \implies A = 1$ AND $B = 1$.
Therefore, the required inputs are A=1, B=1, C=0.